How to solve this question of percentile and percentile groups? The outcome of a standardized test is an integer between 151 and 200, inclusive. The percentiles of 400 test scores are calculated, and the scores are divided into corresponding percentile groups.
Quantity A] The minimum number of integers between 151 and 200 inclusive, that include more than one percentile?
Quantity B] The minimum number of percentile groups that correspond to a score of 200?
Options: 
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A = Quantity B
D) Cannot be determined

Solution as explained :  
Solution part 1
Solution part 2

I did not understand this solution. Kindly help.
 A: This is, in my opinion, a very confusing language the question, and therefore the solution is stated in. I believe once you understand the question well, the answer is almost immediate. Let me ask the question in a clear, but different way.
Say you have 50 baskets, numbered $1,2,\cdots, 50$ and $400$ balls. Say someone comes and distributes the balls in the baskets in a certain fashion. 
Given a distribution of balls, let $M$ be the number of baskets which contain at least 1 percent of the balls. (since one percent of the balls is $400/100=4$, we are counting the number of baskets which end up with $4$ or more balls in them).
Quantity A: If you were free to distribute the balls as you please, how should you distribute them such that $M$ is as little as possible? What is $M$ in this situation? This $M$ is your quantity $A$.
Solution: We first ask: is it possible to achieve $M=0$? Well, no! Because if each basket has $\leq 4$ balls since there are $50$ baskets, the number of balls has to be $\leq 4\times 50=200$. But you have $400$ balls, not $200$.
So next we ask, OK $M=0$ is not possible but is $M=1$ possible? Yes, one easy choice is to put all $400$ balls in just one basket and leave everything else empty. As a result quantity, $A$ is $1$.
Now let's focus on just one basket, say the basket number 50 (the last one). Let $N$ be the percentage of balls this basket contains.
Quantity B: If you were free to distribute the balls as you please how should you distribute them, such that $N$ is as little as possible? What is $N$ in this situation?
Solution: Again we ask: is $N=0$ possible? Well, yes, just don't put any balls in the last basket! So quantity B is zero.
The question and the answer confusingly talk about percentile groups. A percentile group is the number of scores which make up one percent of the total scores. So the number of percentile groups belonging to a basket has is just the percentage it contains. It is rather effectively confusing terminology for this question.
